Universal property characterizing $\Bbb R$

Solution 1:

This question leads directly to the notion of a real closed field. These are ordered fields which "believe that they are $\mathbb{R}$". More precisely, they satisfy the same first order sentences as $\mathbb{R}$. It turns out that already two sentence schemes suffice: Every positive element is a square, and every polynomial of odd degree has a root. There are several other interesting characterizations, which can be found in the Wikipedia article. See also the references given there.

Every ordered field has a real closed algebraic extension, the so called real closure. This gives lots of examples of real closed fields, not isomorphic to $\mathbb{R}$. For example, the real closure of $\mathbb{Q}$ is $\mathbb{R} \cap \overline{\mathbb{Q}}$, i.e. the field of real algebraic numbers.

This strongly suggests that there isn't any first-order universal property of $\mathbb{R}$ within the category of ordered fields, since solutions for universal problems are always isomorphic.

At least we can observe the following: If $K$ is an ordered field, then there is at most one homomorphism of fields $\mathbb{R} \to K$. For a proof, let $f : \mathbb{R} \to K$ be such a homomorphism. Then it has to be monotonic: If $x \leq y$, then $y-x$ is a square, hence also $f(y-x)=f(y)-f(x)$ is a square, which implies $f(x) \leq f(y)$. When $K$ is equipped with the order topology, then $f$ becomes continuous. Since $f|_{\mathbb{Q}}$ is unique and $\mathbb{Q}$ is dense in $\mathbb{R}$, it follows that $f$ is unique.

Observe that $f$ exists if and only if all rational Cauchy sequences have a limit in $K$.